MATH 1A FINAL (7:00 PM VERSION) SOLUTION. (Last edited December 25, 2013 at 9:14pm.)

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MATH A FINAL (7: PM VERSION) SOLUTION (Last edited December 5, 3 at 9:4pm.) Problem. (i) Give the precise defiitio of the defiite itegral usig Riema sums. (ii) Write a epressio for the defiite itegral 5 right had edpoits as sample poits. (iii) Calculate the itegral 5 e e +3 e e +3 d usig ay methods you like. d, givig your aswer as a it ad usig (i) Let f : [a, b] R be a fuctio. For each N, pick a collectio of sample poits,..., so that i (defied as) lies i [a + (i ) b a b a f() d, a + i b a ]. The defiite itegral of f from a to b is i ( ) b a f( i ) provided the it eists ad gives the same value for all possible choices of sample poit. (It is ot assumed that f is cotiuous.) (ii) The required epressio is i e +i( 4 ) e +i( 4 ) + 3 4. (A commo mistake was to write the Riema sum for the case 4.) (iii) Let u e + 3. The du d e. Thus 5 e e + 3 d 5 u(5) u() du u d d u du l(u(5)) l(u()) ( e 5 ) + 3 l e. + 3 by the Substitutio Rule Problem. (i) State carefully ad i full the Fudametal Theorem of Calculus. (ii) Suppose that a particle moves alog the -ais with velocity si( πt ), where t is time i secods. What is the et chage i positio betwee time t secods ad time t secods? (i) Let f be a cotiuous real-valued fuctio o [a, b]. Let g : [a, b] R be defied by g() a f(t) dt. The g is cotiuous o [a, b], differetiable o (a, b), ad g () f(). Furthermore, if F is ay atiderivative of f, the b a f() d F (b) F (a).

MATH A FINAL (7: PM VERSION) SOLUTION (ii) We have Net chage i positio t ( ) πt si dt t ( ( πt π cos ( () π ) )) t t π. Problem 3. (i) Let f be a real-valued fuctio, ad let a ad L be real umbers. What does it mea to say that a + f() L? (ii) State the domai of the fuctio f() 8 5. (iii) Prove, usig the defiitio you gave i part (i), that 5 + 8 5. (i) For all ɛ >, there eists δ > such that if a < < a + δ the f() L < ɛ. (ii) The domai is [5, ). (iii) Let ɛ >. Put δ ɛ 8. Suppose 5 < < 5 + δ. The < 5 < ɛ 8, so < 8 5 < ɛ. Thus Problem 4. 8 5 < ɛ. Thus 5 + 8 5. (i) Let y y(t) be the umber of bacterial cells growig i a culture, where t is time measured i days. Suppose the rate of growth of y(t) is proportioal to y(t). That is, suppose that y (t) Cy(t) where C is costat. Prove carefully that y(t) y()e Ct. (ii) You are give that at time t there are 3 cells i the culture. After 6 days there are 7 cells i the culture. How may days will it take for there to be 8 cells? (i) See Aleader s hadouts. (ii) Let y(t) be the umber of cells i the culture at time t. We have y(t) y()e Ct for some costat C. Problem 5. We have y() 3 ad y(6) 7. Thus 7 3e 6C, which implies C 6 l 9. Let t be the time at which there are 8 cells i culture, i.e. y(t ) 8. The 8 3e Ct implies t l 7 C 6 l 7 l 9 9. (i) A bowl is made by rotatig the regio bouded by the curve y 3, the -ais, ad the lie about the y-ais. Fid the volume of the bowl. (ii) A football is made i the shape of y si, rotated about the -ais betwee ad π. Fid the volume of the football. (You may use the fact that cos(θ) si θ without proof.) (i) The volume of the bowl is (π)( 3 ) d π 4 d π 5 (5 5 ) 64π 5.

MATH A FINAL (7: PM VERSION) SOLUTION 3 (ii) The volume of the football is π π ( ) cos() π(si ) d π π ( ) π si() d π. Problem 6. A ladder that is 3 meters log is leaig agaist a vertical wall ad stadig o horizotal groud. The bottom of the ladder slips. Assume that the top of the ladder stays i cotact with the wall, ad the bottom of the ladder stays i cotact with the groud. Calculate the speed that the bottom of the ladder is movig away from the wall whe the top of the ladder is 5 meters above the groud ad movig dowwards at 3 meters per secod. Let (t) be the distace betwee the bottom of the ladder ad the wall, ad let y(t) be the distace betwee the top of the ladder ad the groud. We have ((t)) + (y(t)) 3 for all t. Let t be the time of iterest. Differetiatig with respect to time gives (t) (t) + y(t)y (t). We have y(t ) 5 (which implies (t ) ) ad y (t ) 3. Thus (t ) y(t)y (t ) (t ) Problem 7. ad y. 5( 3) 5 4. (i) Suppose that y si( 4 ) 5. Fid dy d. You may leave your aswer i terms of both (ii) Suppose y 5. Fid dy d. (i) Differetiatig implicitly gives ad solvig for y gives y + y cos( 4 )(4 3 ) 5 l 5 y 5 l 5 + 4 3 cos( 4 ) y (ii) Takig atural logarithm of both sides gives l(y) 5 l. Differetiatig with respect to gives y y 5 l + 5( ). Thus y 5 (5 l + 5).. Problem 8. Calculate the followig its: e (i) cos(3) (ii) 3 (iii) 7+3 5 4 If you use ay special rules to calculate your aswers, you should state the rule each time you use it.

4 MATH A FINAL (7: PM VERSION) SOLUTION (i) We have e. e (ii) We have (iii) We have cos(3) 3 si(3) 9 cos(3) 9 cos(3 ) 9. 3 7 + 3 6 7 5 4 3 5. 6 Problem 9. Cosider the graph of y 3 3. (i) Give the coordiates of the -itercepts. (ii) Give the coordiates of all critical poits, ad idicate for each if it is a local maimum or local miimum. (iii) Give the coordiates of ay poits of iflectio. (iv) The graph passes through the poit (, 4). Describe the cocavity of the graph at this poit. (i) The -itercepts are the poits o the graph whose y-coordiates are. If (, ) is a -itercept, the 3 3 implies that or 3. Also, (3, ) ad (, ) are poits o the graph. Thus (3, ) ad (, ) are -itercepts. (ii) The critical poits of the graph are where y ; if (, y) is a critical poit, the 3 6. This ca oly happe whe or. Thus the critical poits are (, 4) ad (, ). (iii) The poit (, y) is a poit of iflectio if y. We have y 6, so (, ) is the oly poit of iflectio. (iv) We have y ( ) 6, so the graph is cocave dow. Problem. Let f ad g be real-valued fuctios. Suppose that a f() L ad a g() K. Prove that a (f() + g()) L + K. Let f ad g be fuctios, ad suppose a f() L ad a g() K. Let ɛ >. Sice a f() L, there eists δ > such that if < a < δ, the f() L < ɛ. Also, sice a g() K, there eists δ > such that if < a < δ, the g() K < ɛ. Put δ mi{δ, δ }.

MATH A FINAL (7: PM VERSION) SOLUTION 5 Suppose < a < δ. The f() L < ɛ ad g() K < ɛ. Hece (f() + g()) (L + K) (f() L) + (g() K) f() L + g() K ɛ + ɛ ɛ. Hece a (f() + g()) L + K.